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Math Help - Solving Logarithmic equation

  1. #1
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    Solving Logarithmic equation

    I'm wondering if anyone can help, I'm stuck with the following logarithmic equation:

    x^3*ln(x)-4x*ln(x)=0

    I can easily see why x=1 is a solution (ln(1)=0).Is this the correct procedure for finding the other solutions:

    =(x^3-4x)*ln(x)

    =x(x^2-4)*ln(x)

    =x((x+2)(x-2))*ln(x)

    x= -2 or +2

    The mark scheme for the past paper says the 2 answers are 1 and -2, but I can't see why 2 isn't a valid solution also? Also, is x=0 not a solution because ln(0) is not defined?
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  2. #2
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    Re: Solving Logarithmic equation

    Quote Originally Posted by simon238 View Post
    I'm wondering if anyone can help, I'm stuck with the following logarithmic equation:

    x^3*ln(x)-4x*ln(x)=0

    I can easily see why x=1 is a solution (ln(1)=0).Is this the correct procedure for finding the other solutions:

    =(x^3-4x)*ln(x)

    =x(x^2-4)*ln(x)

    =x((x+2)(x-2))*ln(x)

    x= -2 or +2

    The mark scheme for the past paper says the 2 answers are 1 and -2, but I can't see why 2 isn't a valid solution also? Also, is x=0 not a solution because ln(0) is not defined?
    I think your mark scheme is off because the valid solutions for this are x=1 and x=2.

    You are correct about 0 not being valid because log(0) is not defined and assuming you are solving over the real numbers the same goes for x=-2.
    Thanks from simon238
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  3. #3
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    Re: Solving Logarithmic equation

    Ahh yes, of course, thank you. I guess I just took the mark scheme as gospel, but I couldn't make sense of it
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